# In how many ways can '$6$' things be distributed equally among $2$ groups?

In how many ways can '$6$' things be distributed equally among $2$ groups ?

I tried

$\dbinom{6}{3}\times \dbinom{3}{3}\times 3!\times 3!$

But I am not sure if it is correct .

I look for a short and simple way.

I have studied maths up to $12$th grade.

• Why have you multiplied by 3! and 3! ? – Shailesh Sep 12 '15 at 1:25
• To permute $3$ objects. – R K Sep 12 '15 at 1:28
• actually it was $'3\times 2'$. – R K Sep 12 '15 at 1:31
It's $6\choose 3$ if you have groups $A$ and $B$. You choose three items and give them to $A$; there are $6\choose 3$ ways to do this. You then give the remaining to group $B$.
The $3!s$ are called for if you stipulate the order in which the groups receive the items is important. Otherwise, not.
$${6\choose 3} = 20.$$
If the two groups are indistinguishable and you are just interested in partitioning the items into two parts of size three, we have an overcount of a factor of 2. In that case there are only $10$ ways. You must be careful about exactly what you are counting.